Question

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is:

• A. Zero
• B. $0.5\pi$
• C. $\pi$
• D. $0.707\pi$

Answer 1

The correct option is B $0.5\pi$

$Let$ $y=Asin$ $\omega t$

$\frac{dy}{dx}=A\omega$ $cos\omega t$

$\frac{dy}{dx}=A\omega$ $sin\left(\omega t+\frac{\pi }{2}\right)$

$Acceleration=A{\omega }^{2}sin\omega t$

$The$ $phase$ $difference$ $between$ $acceleration$ $and$ $velocity$ $is$ $\frac{\pi }{2}$ $=0.5\pi$